Question 25

We need to find different sets of integers to make each equation true.

Step 1 — Understanding the first equation

The first equation is $\square + \square \times \square = -36$. Let us call the numbers in the blanks $a$, $b$, and $c$. So, the equation is $a + b \times c = -36$. Remember the order of operations: multiplication happens before addition. This means we first calculate $b \times c$, then add $a$ to that result. So, $a + (b \times c)$ must equal -36.

Step 2 — Finding integers for (a) - Way 1

Let us choose values for $b$ and $c$. We will choose $b$ as -6 and $c$ as 6. First, we multiply $b$ and $c$.

$b \times c = (-6) \times 6$

$= -36$

Now, we need to find $a$ such that $a + (-36) = -36$. To find $a$, we add 36 to both sides.

$a = -36 - (-36)$

$= -36 + 36$

$= 0$

So, the numbers are 0, -6, and 6. Let us check this in the original equation.

$0 + (-6) \times 6 = 0 + (-36)$

$\boxed{0 + (-6) \times 6 = -36}$

Step 3 — Finding integers for (a) - Way 2

Let us choose $b$ as -5 and $c$ as 8. First, we multiply $b$ and $c$.

$b \times c = (-5) \times 8$

$= -40$

Now, we need to find $a$ such that $a + (-40) = -36$. To find $a$, we add 40 to both sides.

$a = -36 - (-40)$

$= -36 + 40$

$= 4$

So, the numbers are 4, -5, and 8. Let us check this in the original equation.

$4 + (-5) \times 8 = 4 + (-40)$

$\boxed{4 + (-5) \times 8 = -36}$

Step 4 — Finding integers for (a) - Way 3

Let us choose $b$ as -8 and $c$ as 4. First, we multiply $b$ and $c$.

$b \times c = (-8) \times 4$

$= -32$

Now, we need to find $a$ such that $a + (-32) = -36$. To find $a$, we add 32 to both sides.

$a = -36 - (-32)$

$= -36 + 32$

$= -4$

So, the numbers are -4, -8, and 4. Let us check this in the original equation.

$-4 + (-8) \times 4 = -4 + (-32)$

$\boxed{-4 + (-8) \times 4 = -36}$

Step 5 — Finding integers for (a) - Way 4

Let us choose $b$ as -8 and $c$ as 6. First, we multiply $b$ and $c$.

$b \times c = (-8) \times 6$

$= -48$

Now, we need to find $a$ such that $a + (-48) = -36$. To find $a$, we add 48 to both sides.

$a = -36 - (-48)$

$= -36 + 48$

$= 12$

So, the numbers are 12, -8, and 6. Let us check this in the original equation.

$12 + (-8) \times 6 = 12 + (-48)$

$\boxed{12 + (-8) \times 6 = -36}$

Step 6 — Finding integers for (a) - Way 5

Let us choose $b$ as 3 and $c$ as -11. First, we multiply $b$ and $c$.

$b \times c = 3 \times (-11)$

$= -33$

Now, we need to find $a$ such that $a + (-33) = -36$. To find $a$, we add 33 to both sides.

$a = -36 - (-33)$

$= -36 + 33$

$= -3$

So, the numbers are -3, 3, and -11. Let us check this in the original equation.

$-3 + 3 \times (-11) = -3 + (-33)$

$\boxed{-3 + 3 \times (-11) = -36}$

Step 7 — Understanding the second equation

The second equation is $(\square - \square) \times \square = 12$. Let us call the numbers in the blanks $x$, $y$, and $z$. So, the equation is $(x - y) \times z = 12$. Remember the order of operations: we first calculate inside the parenthesis $(x - y)$. Then we multiply that result by $z$. The product of $(x - y)$ and $z$ must be 12. This means $z$ must be a factor of 12.

Step 8 — Finding integers for (b) - Way 1

Let us choose a value for $z$. We will choose $z$ as 1. Now, we need $(x - y) \times 1 = 12$. This means $(x - y)$ must be 12. We need two numbers $x$ and $y$ whose difference is 12. Let us choose $y$ as 1. Then $x - 1 = 12$. To find $x$, we add 1 to both sides.

$x = 12 + 1$

$= 13$

So, the numbers are 13, 1, and 1. Let us check this in the original equation.

$(13 - 1) \times 1 = 12 \times 1$

$\boxed{(13 - 1) \times 1 = 12}$

Step 9 — Finding integers for (b) - Way 2

Let us choose $z$ as 2. Now, we need $(x - y) \times 2 = 12$. To find $(x - y)$, we divide 12 by 2.

$x - y = 12 \div 2$

$= 6$

We need two numbers $x$ and $y$ whose difference is 6. Let us choose $y$ as 4. Then $x - 4 = 6$. To find $x$, we add 4 to both sides.

$x = 6 + 4$

$= 10$

So, the numbers are 10, 4, and 2. Let us check this in the original equation.

$(10 - 4) \times 2 = 6 \times 2$

$\boxed{(10 - 4) \times 2 = 12}$

Step 10 — Finding integers for (b) - Way 3

Let us choose $z$ as -6. Now, we need $(x - y) \times (-6) = 12$. To find $(x - y)$, we divide 12 by -6.

$x - y = 12 \div (-6)$

$= -2$

We need two numbers $x$ and $y$ whose difference is -2. Let us choose $y$ as 3. Then $x - 3 = -2$. To find $x$, we add 3 to both sides.

$x = -2 + 3$

$= 1$

So, the numbers are 1, 3, and -6. Let us check this in the original equation.

$(1 - 3) \times (-6) = (-2) \times (-6)$

$\boxed{(1 - 3) \times (-6) = 12}$

Step 11 — Finding integers for (b) - Way 4

Let us choose $z$ as 3. Now, we need $(x - y) \times 3 = 12$. To find $(x - y)$, we divide 12 by 3.

$x - y = 12 \div 3$

$= 4$

We need two numbers $x$ and $y$ whose difference is 4. Let us choose $y$ as 10. Then $x - 10 = 4$. To find $x$, we add 10 to both sides.

$x = 4 + 10$

$= 14$

So, the numbers are 14, 10, and 3. Let us check this in the original equation.

$(14 - 10) \times 3 = 4 \times 3$

$\boxed{(14 - 10) \times 3 = 12}$

Step 12 — Finding integers for (b) - Way 5

Let us choose $z$ as 4. Now, we need $(x - y) \times 4 = 12$. To find $(x - y)$, we divide 12 by 4.

$x - y = 12 \div 4$

$= 3$

We need two numbers $x$ and $y$ whose difference is 3. Let us choose $y$ as 13. Then $x - 13 = 3$. To find $x$, we add 13 to both sides.

$x = 3 + 13$

$= 16$

So, the numbers are 16, 13, and 4. Let us check this in the original equation.

$(16 - 13) \times 4 = 3 \times 4$

$\boxed{(16 - 13) \times 4 = 12}$

Step 13 — Understanding the third equation

The third equation is $(\square - (\square - \square)) = -1$. Let us call the numbers in the blanks $p$, $q$, and $r$. So, the equation is $(p - (q - r)) = -1$. Remember the order of operations: we always work from the innermost parenthesis first. First, we calculate $(q - r)$. Then we subtract this result from $p$. The final result must be -1.

Step 14 — Finding integers for (c) - Way 1

Let us choose values for $q$ and $r$. We will choose $q$ as 10 and $r$ as 4. First, we calculate $(q - r)$.

$q - r = 10 - 4$

$= 6$

Now, we need to find $p$ such that $p - 6 = -1$. To find $p$, we add 6 to both sides.

$p = -1 + 6$

$= 5$

So, the numbers are 5, 10, and 4. Let us check this in the original equation.

$(5 - (10 - 4)) = 5 - 6$

$\boxed{(5 - (10 - 4)) = -1}$

Step 15 — Finding integers for (c) - Way 2

Let us choose $q$ as 3 and $r$ as 2. First, we calculate $(q - r)$.

$q - r = 3 - 2$

$= 1$

Now, we need to find $p$ such that $p - 1 = -1$. To find $p$, we add 1 to both sides.

$p = -1 + 1$

$= 0$

So, the numbers are 0, 3, and 2. Let us check this in the original equation.

$(0 - (3 - 2)) = 0 - 1$

$\boxed{(0 - (3 - 2)) = -1}$

Step 16 — Finding integers for (c) - Way 3

Let us choose $q$ as 1 and $r$ as 1. First, we calculate $(q - r)$.

$q - r = 1 - 1$

$= 0$

Now, we need to find $p$ such that $p - 0 = -1$. This means $p$ must be -1.

$p = -1$

So, the numbers are -1, 1, and 1. Let us check this in the original equation.

$(-1 - (1 - 1)) = -1 - 0$

$\boxed{(-1 - (1 - 1)) = -1}$

Step 17 — Finding integers for (c) - Way 4

Let us choose $q$ as 0 and $r$ as 4. First, we calculate $(q - r)$.

$q - r = 0 - 4$

$= -4$

Now, we need to find $p$ such that $p - (-4) = -1$. Remember that subtracting a negative number is the same as adding a positive number. So, $p + 4 = -1$. To find $p$, we subtract 4 from both sides.

$p = -1 - 4$

$= -5$

So, the numbers are -5, 0, and 4. Let us check this in the original equation.

$(-5 - (0 - 4)) = -5 - (-4)$

$= -5 + 4$

$\boxed{(-5 - (0 - 4)) = -1}$

Step 18 — Finding integers for (c) - Way 5

Let us choose $q$ as -5 and $r$ as 4. First, we calculate $(q - r)$.

$q - r = -5 - 4$

$= -9$

Now, we need to find $p$ such that $p - (-9) = -1$. Remember that subtracting a negative number is the same as adding a positive number. So, $p + 9 = -1$. To find $p$, we subtract 9 from both sides.

$p = -1 - 9$

$= -10$

So, the numbers are -10, -5, and 4. Let us check this in the original equation.

$(-10 - (-5 - 4)) = -10 - (-9)$

$= -10 + 9$

$\boxed{(-10 - (-5 - 4)) = -1}$

Answer

(a) [ ] + [ ] x [ ] = -36 (i) 0 + (-6) x 6 = -36 (ii) 4 + (-5) x 8 = -36 (iii) -4 + (-8) x 4 = -36 (iv) 12 + (-8) x 6 = -36 (v) -3 + 3 x (-11) = -36

(b) ([ ] - [ ]) x [ ] = 12 (i) (13 - 1) x 1 = 12 (ii) (10 - 4) x 2 = 12 (iii) (1 - 3) x (-6) = 12 (iv) (14 - 10) x 3 = 12 (v) (16 - 13) x 4 = 12

(c) ([ ] - ([ ] - [ ])) = -1 (i) (5 - (10 - 4)) = -1 (ii) (0 - (3 - 2)) = -1 (iii) (-1 - (1 - 1)) = -1 (iv) (-5 - (0 - 4)) = -1 (v) (-10 - (-5 - 4)) = -1